Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.
Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !
vf = (initial velocity) + (acceleration)*(time)
vf = (3 m/s) + (5 m/s²)*(4 sec)
vf = (3 m/s) + (20 m/s)
vf = 23 m/s
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t).
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
Answer:
The mass of the second person is 28.91 kg
Explanation:
Given;
mass of the first adult, m₁ = 54.2 kg
distance of the first adult from the point of balance, x = 1.20 m
mass of the second adult, = m₂
distance of the second adult from the point of balance, y = 2.25 m
Taking moment about the point of balance, we will have
m₁x = m₂y
54.2 x 1.2 = 2.25y
2.25y = 65.04
y = 65.04/2.25
y = 28.91 kg
Therefore, the mass of the second person is 28.91 kg
Potential energy due to gravity = Ep = mgh [symbols have their usual meaning ]
Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.
